Question

A small ring is given some velocity along the axis of a solenoid and it remains coaxial with solenoid. Current in solenoid is \(i = 10 \sin(\omega t)\); \(\omega = 1000\) rad/s. Number of turns per unit length is 500/m. Radius of ring is 1 cm and its resistance is 10\(\Omega\). Find RMS value of induced current in the ring :

• A. \(\sqrt{2} \times 10^{-5}\) A
• B. \(3 \times 10^{-4}\) A
• C. \(\sqrt{2} \times 10^{-4}\) A
• D. \(5 \times 10^{-6}\) A

Hint:

The velocity of the ring does not affect the induced EMF in this case because the magnetic field inside an ideal solenoid is uniform. The EMF is purely due to the time-varying current (transformer EMF). If the field were non-uniform, there would also be a motional EMF component.

Answer 1

The Correct Option is C

To determine the RMS value of the induced current in the ring, we need to follow a series of logical steps and calculations involving electromagnetic induction concepts. 

1. The solenoid has a time-varying current \( i(t) = 10 \sin(\omega t) \) where \( \omega = 1000 \) rad/s.
2. The number of turns per unit length of the solenoid is given as \( n = 500 \, \text{turns/m} \).
3. The radius of the ring is \( r = 1 \, \text{cm} = 0.01 \, \text{m} \) and its resistance is \( R = 10 \, \Omega \).
4. We use Faraday's Law of electromagnetic induction to find the induced EMF \( \mathcal{E} \) in the ring. The magnetic field inside a solenoid is given by \( B = \mu_0 n i \), where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T} \cdot \text{m/A} \) is the permeability of free space.
5. The magnetic flux through the ring is \( \Phi = B \cdot \pi r^2 = \mu_0 n i \pi r^2 \).
6. The induced EMF in the ring by Faraday's Law is: 

\[\mathcal{E} = -\frac{d\Phi}{dt} = -\pi r^2 \mu_0 n \frac{di}{dt}\]

1. Substitute \( i = 10 \sin(\omega t) \), then: 

\[\frac{di}{dt} = 10\omega \cos(\omega t)\]

1. Thus, the induced EMF becomes: 

\[\mathcal{E} = -\pi r^2 \mu_0 n \times 10\omega \cos(\omega t)\]

1. The magnitude of the instantaneous EMF is: 

\[\mathcal{E}_0 = \pi (0.01)^2 \times 4\pi \times 10^{-7} \times 500 \times 1000 \times 10\]

Calculate the above expression to find \( \mathcal{E}_0 \): 

\[\mathcal{E}_0 = \pi \cdot 0.0001 \cdot 4\pi \cdot 10^{-7} \cdot 500 \cdot 10^4 \approx 2 \times 10^{-3} \, \text{V}\]

1. The RMS value of the current \( I_{\text{rms}} = \frac{\mathcal{E}_0}{\sqrt{2} R} \) due to the sinusoidal nature of the induced EMF. So: 

\[I_{\text{rms}} = \frac{2 \times 10^{-3}}{\sqrt{2} \cdot 10} = \sqrt{2} \times 10^{-4} \, \text{A}\]

Thus, the RMS value of the induced current in the ring is \( \sqrt{2} \times 10^{-4} \, \text{A} \), which matches the correct answer.